Question

how many terms of the AP 2, 4, 6, 8, 10 are needed to get some 210

Answer 1

Your question needs a correction.

Correct question : - how many terms of the AP 2, 4, 6, 8, 10 are needed to get sum 210.

       Solution : -

Let there are n terms.

Given, 1st term = a = 2

           2nd term = a₂ = 4

           3rd term = a₃ = 6

∴ Common Difference = a₃ - a₂

                                       = 6 - 4

                                       = 2

we know that the sum of n terms of AS is $\dfrac{n}{2}{2a+(n-1)d]$.

Substituting the given values in the formula given above.

210 = $\dfrac{n}{2}[2(2)+(n-1)(2)$

210 = $\dfrac{n}{2}[4+2n-2]$

210 = $\dfrac{n}{2}[2n+2]$

210 = $\dfrac{n}{2}[ 2(n+1)]$

210 = n( n + 1  )

210 = n^2 + n

n^2 + n - 210 = 0

n^2 + ( 15 - 14 )n - 210 = 0

n^2 + 15n - 14n - 210 = 0

n( n + 15 ) - 14( n + 15 ) = 0  

( n + 15 ) ( n - 14 )= 0

n = - 15 or n = 14

As n is the number of terms of AP, it can't be negative. Therefore, n = 14

Hence there are 14 terms in the AS.

Answer 2

In the attachment I have answered this problem.

Concept:

Sum of n terms of an AP

= n/2[2a+(n-1)d]

From the attachment it is clear that 14 terms are needed to get the sum 210.

See the attachment for detailed solution.